So we all know (and love) the typical dynamic equation of a robotic arm:

$$ M(q) \ddot{q}+C(q,\dot{q})\dot{q}+G(q)=\tau $$

Where M is the mass matrix, G is gravity and C is Coriolis and Centrifugal forces.

Coriolis and Centrifugal forces are pseudo forces that are experienced only for the frame that is attached to the body that is rotating. But my understanding is that all calculations happen relative to the base frame, which is static. The need for Coriolis and Centrifugal forces would imply that the forces are actually expressed in the frame that is attached to each link.

To put my question simply, why do we need the coriolis and centrifugal forces in the dynamic equation of a robotic arm (since everything is done relative to the initial/base frame)?

(p.s. feel free to use a single or double pendulum robotic arm example if you would prefer).

  • $\begingroup$ is this a school assignment question? $\endgroup$
    – jsotola
    Sep 20, 2019 at 15:05
  • $\begingroup$ @jsotola no. just me trying to understand better. $\endgroup$
    – Metalzero2
    Sep 20, 2019 at 15:05

1 Answer 1


But my understanding is that all calculations happen relative to the base frame

This is incorrect. Take a look at this two-link arm manipulator. The point ($x_2,y_2$) is obviously expressed in a moving frame $y_1,x_1$. Another way to look at the problem is to use Lagrangian approach which free you from constraint forces.

enter image description here

  • $\begingroup$ Thanks for you answer. I think I get what you mean. But what about the case where the robot had only one link (so the above example with l1 only)? Would you not need to use coriolis and centrifugal forces? $\endgroup$
    – Metalzero2
    Sep 21, 2019 at 11:45
  • $\begingroup$ What do you think? $\endgroup$
    – CroCo
    Sep 21, 2019 at 16:36
  • $\begingroup$ Based on the fact that in your answer you referred only to the second link because its coordinates are expressed based on the rotating frame (x1-y1), I would say no, those forces are not needed, since frame (x0-y0) is not moving. $\endgroup$
    – Metalzero2
    Sep 21, 2019 at 18:19
  • $\begingroup$ The lagrange does not free one from constraint forces...do you mean you can use it to reduce the system and ‘hide them’ ? $\endgroup$ Sep 21, 2019 at 18:57
  • $\begingroup$ @DimitrisPantelis true $\endgroup$
    – CroCo
    Sep 23, 2019 at 21:52

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